Comparing energy transfers using work-energy principles

Steps:

• Calculate kinetic energy lost in A: $\frac{1}{2}mv^2 = \frac{1}{2} \times 80 \times 2^2 = 160$ J.
• Calculate gravitational potential energy in B: $mgh = 80 \times 9.8 \times 2 \approx 1568$ J.
• Calculate work done in C: $F \times d = 70 \times 2 = 140$ J.
• Calculate electrical energy in D: $P \times t = 70 \times 20 = 1400$ J.

Why C is correct:

• It involves the smallest work transfer, $W = Fd = 140$ J, per the work-energy theorem.

Why the others are wrong:

• A transfers 160 J of kinetic energy, more than C.
• B transfers about 1568 J of potential energy, much more than C.
• D transfers 1400 J of electrical energy, more than C.

Final answer: C